Recap general linear model

1 Breast cancer example

• part of study https://doi.org/10.1093/jnci/djj052)
• Histologic grade in breast cancer clinically prognostic. Association of histologic grade on expression of KPNA2 gene that is known to be associated with poor BC prognosis.
• Population: all current and future breast cancer patients

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2 Data Exploration

2.1 Import

library(tidyverse)
gene <- read.table("https://raw.githubusercontent.com/statOmics/SGA2020/data/gse2990BreastcancerOneGene.txt",header=TRUE)
head(gene)

##    sample_name grade node size age     gene
## 28   OXFT_2221     3    1  5.5  76 367.8179
## 29    OXFT_209     3    1  2.5  66 590.3576
## 30   OXFT_1769     1    1  3.5  86 346.6583
## 31    OXFT_928     1    0  1.1  47 118.6996
## 32   OXFT_2093     1    1  2.2  74 519.4489
## 33   OXFT_1770     1    1  1.7  69 258.4455

We will transform the variable grade and node to a factor

gene$grade <- as.factor(gene$grade)
gene$node <- as.factor(gene$node)

2.2 Summary statistics

geneSum <- gene %>%
  group_by(grade) %>%
  summarize(mean = mean(gene),
            sd = sd(gene),
            n=length(gene)
            ) %>%
  mutate(se = sd/sqrt(n))
geneSum

## # A tibble: 2 x 5
##   grade  mean    sd     n    se
##   <fct> <dbl> <dbl> <int> <dbl>
## 1 1      264.  117.    19  26.7
## 2 3      606.  267.    17  64.9

2.3 Visualisation

gene %>%
  ggplot(aes(x=grade,y=gene)) +
  geom_boxplot(outlier.shape=NA) +
  geom_point()

We can also save the plots as objects for later use!

p1 <- gene %>%
  ggplot(aes(x=grade,y=gene)) +
  geom_boxplot(outlier.shape=NA) +
  geom_point()

p2 <- gene %>%
  filter(grade==1) %>%
  ggplot(aes(sample=gene)) +
  geom_qq() +
  geom_qq_line()

p3 <- gene %>%
  filter(grade==1) %>%
  ggplot(aes(sample=gene)) +
  geom_qq() +
  geom_qq_line()

p1

p2

p3

2.4 Research questions

Researchers want to assess the association of the histolical grade on KPNA2 gene expression

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2.5 Estimation of effect size and standard error

effectSize <- tibble(
  delta = geneSum$mean[2]- geneSum$mean[1],
  seDelta = geneSum %>%
    pull(se) %>%
    .^2 %>%
    sum %>%
    sqrt
  )
effectSize

## # A tibble: 1 x 2
##   delta seDelta
##   <dbl>   <dbl>
## 1  342.    70.2

3 Statistical Inference

• Researchers want to assess the association of histological grade on KPNA2 gene expression
• Inference?

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• Researchers want to assess the association of histological grade on KPNA2 gene expression
• Inference?
• testing + CI $ $ Assumptions

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• In general we start from alternative hypothese \(H_A\): we want to show an association

• Gene expression of grade 1 and grade 3 patients is on average different

• But, we will assess it by falsifying the opposite:

• The average KPNA2 gene expression of grade 1 and grade 3 patients is equal

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• How likely is it to observe an equal or more extreme association than the one observed in the sample when the null hypothesis is true?

• When we make assumptions about the distribution of our test statistic we can quantify this probability: p-value.

• If the p-value is below a significance threshold \(\alpha\) we reject the null hypothesis

We control the probability on a false positive result at the \(\alpha\)-level (type I error)

• The p-value will only be calculated correctly if the underlying assumptions hold!

library(gridExtra)
p1

grid.arrange(p2,p3,ncol=2)

t.test(gene~grade,data=gene)

## 
##  Welch Two Sample t-test
## 
## data:  gene by grade
## t = -4.8806, df = 21.352, p-value = 7.61e-05
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
##  -488.1734 -196.6587
## sample estimates:
## mean in group 1 mean in group 3 
##        263.5516        605.9677

effectSize <- effectSize %>%
  mutate(t.stat=delta/seDelta) %>%
  mutate(p.value= pt(-abs(t.stat),21.352)*2)

effectSize

## # A tibble: 1 x 4
##   delta seDelta t.stat   p.value
##   <dbl>   <dbl>  <dbl>     <dbl>
## 1  342.    70.2   4.88 0.0000761

• Intensities are often not normally distributed and have a mean variance relation
• Commonly log2-transformed
• Differences on log scale:

\[ \log_2(B) - \log_2(A) = \log_2 \frac{B}{A} = \log_2 FC_{\frac{B}{A}} \]

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3.1 Log transformation

gene <- gene %>%
  mutate(lgene = log2(gene))

p1 <- gene %>%
  ggplot(aes(x=grade,y=lgene)) +
  geom_boxplot(outlier.shape=NA) +
  geom_point()

p2 <- gene %>%
  filter(grade==1) %>%
  ggplot(aes(sample=lgene)) +
  geom_qq() +
  geom_qq_line()

p3 <- gene %>%
  filter(grade==1) %>%
  ggplot(aes(sample=lgene)) +
  geom_qq() +
  geom_qq_line()

p1

grid.arrange(p2,p3,ncol=2)

logtest <- t.test(lgene~grade,data=gene)
logtest

## 
##  Welch Two Sample t-test
## 
## data:  lgene by grade
## t = -6.0508, df = 33.962, p-value = 7.432e-07
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
##  -1.6236927 -0.8072052
## sample estimates:
## mean in group 1 mean in group 3 
##        7.912963        9.128412

log2FC <- logtest$estimate[2]-logtest$estimate[1]
log2FC

## mean in group 3 
##        1.215449

names(log2FC) <- "g3-g1"
2^log2FC

##   g3-g1 
## 2.32213

3.2 Conclusion

There is a extremely significant association of the histological grade on the gene expression in tumor tissue. On average, the gene expression for the grade 3 patients is 2.3221304 times higher than the gene expression in grade 1 patients (95% CI [1.75, 3.08], \(p<<0.001\)).

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The patients also differ in the their lymph node status. Hence, we have a two factorial design: grade x lymph node status!!!

Solution??

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4 General Linear Model

How can we integrate multiple factors and continuous covariates in linear model.

\[ y_i= \beta_0 + \beta_1 x_{i,1} + \beta_2 x_{i,2} + \beta_{12}x_{i,1}x_{i,2}+\epsilon_i, \] with

• \(x_{i,1}\) a dummy variable for histological grade: \(x_{i,1}=\begin{cases} 0& \text{grade 1}\\ 1& \text{grade 3} \end{cases}\)
• \(x_{i,2}\) a dummy variable for : \(x_{i,2}=\begin{cases} 0& \text{lymph nodes were not removed}\\ 1& \text{lymph nodes were removed} \end{cases}\)
• \(\epsilon_i\)?

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4.1 Implementation in R

lm1 <- lm(gene~grade*node,data=gene)
summary(lm1)

## 
## Call:
## lm(formula = gene ~ grade * node, data = gene)
## 
## Residuals:
##     Min      1Q  Median      3Q     Max 
## -356.85  -91.98  -31.47   53.00  612.73 
## 
## Coefficients:
##              Estimate Std. Error t value Pr(>|t|)    
## (Intercept)    207.60      60.00   3.460  0.00155 ** 
## grade3         434.21      84.85   5.117 1.41e-05 ***
## node1          132.88      92.46   1.437  0.16040    
## grade3:node1  -234.43     136.92  -1.712  0.09655 .  
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 199 on 32 degrees of freedom
## Multiple R-squared:  0.4809, Adjusted R-squared:  0.4322 
## F-statistic: 9.881 on 3 and 32 DF,  p-value: 9.181e-05

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4.2 Assumptions

plot(lm1)

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4.3 Breast cancer example

• Paper: https://doi.org/10.1093/jnci/djj052
• Histologic grade in breast cancer provides clinically important prognostic information. Two factors have to be concidered: Histologic grade (grade 1 and grade 3) and lymph node status (0 vs 1). The researchers assessed gene expression of the KPNA2 gene a protein-coding gene associated with breast cancer and are mainly interested in the association of histological grade. Note, that the gene variable consists of background corrected normalized intensities obtained with a microarray platform. Upon log-transformation, they are known to be a good proxy for the \(\log\) transformed concentration of gene expression product of the KPNA2 gene.
• Research questions and translate them towards model parameters (contrasts)?
• Make an R markdown file to answer the research questions

library(ExploreModelMatrix)
explMx <- VisualizeDesign(gene,designFormula = ~grade*node)
explMx$plotlist

## [[1]]

You can also explore the model matrix interactively:

ExploreModelMatrix(gene,designFormula = ~grade*node)

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5 Linear regression in matrix form

5.1 Scalar form

• Consider a vector of predictors \(\mathbf{x}=(x_1,\ldots,x_p)\) and
• a real-valued response \(Y\)
• then the linear regression model can be written as \[ Y=f(\mathbf{x}) +\epsilon=\beta_0+\sum\limits_{j=1}^p x_j\beta_j + \epsilon \] with i.i.d. \(\epsilon\sim N(0,\sigma^2)\)

5.2 Matrix form

• \(n\) observations \((\mathbf{x}_1,y_1) \ldots (\mathbf{x}_n,y_n)\)
• Regression in matrix notation \[\mathbf{Y}=\mathbf{X\beta} + \mathbf{\epsilon}\] with \(\mathbf{Y}=\left[\begin{array}{c}y_1\\ \vdots\\y_n\end{array}\right]\), \(\mathbf{X}=\left[\begin{array}{cccc} 1&x_{11}&\ldots&x_{1p}\\ \vdots&\vdots&&\vdots\\ 1&x_{n1}&\ldots&x_{np} \end{array}\right]\), \(\mathbf{\beta}=\left[\begin{array}{c}\beta_0\\ \vdots\\ \beta_p\end{array}\right]\) and \(\mathbf{\epsilon}=\left[\begin{array}{c} \epsilon_1 \\ \vdots \\ \epsilon_n\end{array}\right]\)

5.3 Least Squares (LS)

• Minimize the residual sum of squares \[\begin{eqnarray*} RSS(\mathbf{\beta})&=&\sum\limits_{i=1}^n e^2_i\\ &=&\sum\limits_{i=1}^n \left(y_i-\beta_0-\sum\limits_{j=1}^p x_{ij}\beta_j\right)^2 \end{eqnarray*}\]
• or in matrix notation \[\begin{eqnarray*} RSS(\mathbf{\beta})&=&(\mathbf{Y}-\mathbf{X\beta})^T(\mathbf{Y}-\mathbf{X\beta})\\ &=&\Vert \mathbf{Y}-\mathbf{X\beta}\Vert^2 \end{eqnarray*}\] with the \(L_2\)-norm of a \(p\)-dim. vector \(v\) \(\Vert \mathbf{v} \Vert=\sqrt{v_1^2+\ldots+v_p^2}\) \(\rightarrow\)

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5.3.1 Minimize RSS

\[ \begin{array}{ccc} \frac{\partial RSS}{\partial \mathbf{\beta}}&=&\mathbf{0}\\\\ \frac{(\mathbf{Y}-\mathbf{X\beta})^T(\mathbf{Y}-\mathbf{X\beta})}{\partial \mathbf{\beta}}&=&\mathbf{0}\\\\ -2\mathbf{X}^T(\mathbf{Y}-\mathbf{X\beta})&=&\mathbf{0}\\\\ \mathbf{X}^T\mathbf{X\beta}&=&\mathbf{X}^T\mathbf{Y}\\\\ \hat{\mathbf{\beta}}&=&(\mathbf{X}^T\mathbf{X})^{-1}\mathbf{X}^T\mathbf{Y} \end{array} \]

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5.4 Variance Estimator?

\[ \begin{array}{ccl} \hat{\boldmath{\Sigma}}_{\hat{\mathbf{\beta}}} &=&\text{var}\left[(\mathbf{X}^T\mathbf{X})^{-1}\mathbf{X}^T\mathbf{Y}\right]\\\\ &=&(\mathbf{X}^T\mathbf{X})^{-1}\mathbf{X}^T\text{var}\left[\mathbf{Y}\right]\mathbf{X}(\mathbf{X}^T\mathbf{X})^{-1}\\\\ &=&(\mathbf{X}^T\mathbf{X})^{-1}\mathbf{X}^T(\mathbf{I}\sigma^2)\mathbf{X}(\mathbf{X}^T\mathbf{X})^{-1} \\\\ &=&(\mathbf{X}^T\mathbf{X})^{-1}\mathbf{X}^T\mathbf{I}\quad\mathbf{X}(\mathbf{X}^T\mathbf{X})^{-1}\sigma^2\\\\ %\hat{\boldmath{\Sigma}}_{\hat{\mathbf{\beta}}}&=&(\mathbf{X}^T\mathbf{X})^{-1}\mathbf{X}^T\var\left[\mathbf{Y}\right](\mathbf{X}^T\mathbf{X})^{-1}\mathbf{X}\\ &=&(\mathbf{X}^T\mathbf{X})^{-1}\mathbf{X}^T\mathbf{X}(\mathbf{X}^T\mathbf{X})^{-1}\sigma^2\\\\ &=&(\mathbf{X}^T\mathbf{X})^{-1}\sigma^2 \end{array} \]

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6 Homework: Adopt the gene analysis on log scale in matrix form!

1. Study the solution of the exercise to understand the analysis in R https://gtpb.github.io/PSLS20/pages/08-multipleRegression/08-multipleRegression_KPNA2.html

2. Calculate

• model parameters and contrasts of interest
• standard errors, standard errors on contrasts
• t-test statistics on the model parameters and contrasts of interest

3. Compare your results with the output of the lm(.) function

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6.1 Inspiration

Tip: details on the implementation can be found in the book of Faraway (chapter 2). https://people.bath.ac.uk/jjf23/book/

• Design matrix

X <- model.matrix(~grade*node,data=gene)

• Transpose of a matrix: use function t(.)

t(X)

##              1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23
## (Intercept)  1 1 1 1 1 1 1 1 1  1  1  1  1  1  1  1  1  1  1  1  1  1  1
## grade3       1 1 0 0 0 0 0 1 0  1  0  1  0  1  1  0  1  0  0  0  0  1  0
## node1        1 1 1 0 1 1 0 0 0  1  1  0  0  0  0  0  0  0  0  0  0  1  1
## grade3:node1 1 1 0 0 0 0 0 0 0  1  0  0  0  0  0  0  0  0  0  0  0  1  0
##              24 25 26 27 28 29 30 31 32 33 34 35 36
## (Intercept)   1  1  1  1  1  1  1  1  1  1  1  1  1
## grade3        1  1  0  1  1  1  1  0  1  0  0  1  0
## node1         0  0  1  0  1  0  0  1  0  0  1  1  0
## grade3:node1  0  0  0  0  1  0  0  0  0  0  0  1  0
## attr(,"assign")
## [1] 0 1 2 3
## attr(,"contrasts")
## attr(,"contrasts")$grade
## [1] "contr.treatment"
## 
## attr(,"contrasts")$node
## [1] "contr.treatment"

• Matrix product %*% operator

t(X)%*%X

##              (Intercept) grade3 node1 grade3:node1
## (Intercept)           36     17    14            6
## grade3                17     17     6            6
## node1                 14      6    14            6
## grade3:node1           6      6     6            6

• Degrees of freedom of a model? $$ df = n-p$

summary(lm1)

## 
## Call:
## lm(formula = gene ~ grade * node, data = gene)
## 
## Residuals:
##     Min      1Q  Median      3Q     Max 
## -356.85  -91.98  -31.47   53.00  612.73 
## 
## Coefficients:
##              Estimate Std. Error t value Pr(>|t|)    
## (Intercept)    207.60      60.00   3.460  0.00155 ** 
## grade3         434.21      84.85   5.117 1.41e-05 ***
## node1          132.88      92.46   1.437  0.16040    
## grade3:node1  -234.43     136.92  -1.712  0.09655 .  
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 199 on 32 degrees of freedom
## Multiple R-squared:  0.4809, Adjusted R-squared:  0.4322 
## F-statistic: 9.881 on 3 and 32 DF,  p-value: 9.181e-05

dfRes <- (nrow(X)-ncol(X))
dfRes

## [1] 32

• Variance estimator: MSE

\[ \hat \sigma^2 = \sum\limits_{i=1}^n \epsilon_i^2}{n-p} \]

• Invert matrix: use function solve(.)

• Diagonal elements of a matrix: use function diag(.)

t(X)%*%X

##              (Intercept) grade3 node1 grade3:node1
## (Intercept)           36     17    14            6
## grade3                17     17     6            6
## node1                 14      6    14            6
## grade3:node1           6      6     6            6

diag(t(X)%*%X)

##  (Intercept)       grade3        node1 grade3:node1 
##           36           17           14            6